Question

In a trapezium PQRS, as shown in the figure, $PQ||SR||TU.IfTP=\frac{1}{2}PS,$ then the ar(PQRS) is equal to

• A.
  $4\left[ar\right(STUR)-ar(\Delta VSR\left)\right]$
• B.
  $2ar\left(PQUT\right)$
• C.
  $2ar\left(\Delta SRQ\right)$
• D.
  $2ar\left(\Delta QPS\right)$

Answer 1

The correct option is A

$4\left[ar\right(STUR)-ar(\Delta VSR\left)\right]$
Given that PQRS is a trapezium such that $TP=\frac{1}{2}PSandPQ||SR||TU.$

In $\Delta SPN,$ by the converse of mid-point theorem,

$SO=ON(\because TO||PN)$

Consider: $4\left[Ar\right(STUR)–Ar(\Delta VSR\left)\right]=4[(\frac{1}{2}\times (TU+RS)\times \frac{SN}{2})-(\frac{1}{2}\times RS\times \frac{SN}{2})]=4(\frac{1}{2}\times TU\times \frac{SN}{2})=TU\times SN$

Now, T is the mid-point of SP.

$\therefore In\Delta SPQ,$ by the converse of mid-point theorem,

$SV=VQ(\because TV||PQ)\text{Thus, V is the mid-point of SQ,}Similarly,in\Delta SRQ,\text{by the converse of mid-point theorem,}RU=UQ(\because VU||SR)$

Thus, U is the mid-point of RQ.

Also, $TV=\frac{1}{2}PQandVU=\frac{1}{2}SR\therefore TV+VU=\frac{1}{2}\times (PQ+SR)\Rightarrow TU=\frac{1}{2}\times (PQ+SR)...\left(ii\right)\therefore 4\left[Ar\right(STUR)-Ar(\Delta VSR\left)\right]=TU\times SN\left[From\right(i\left)\right]=\frac{1}{2}\times (PQ+SR)\times SN\text{[From (i) and (ii) ]}=Ar\left(PQRS\right)$

Hence, the correct answer is option (a).